Drawing that in a coordinate-system, you might notice that the point C has to lie on the function f(x)=1.
The altitude to the side AC equals 3 units (=1-(-2)).
Hence (using area formula: h(c)*c/2) we get:

5=c*3/2 $\ \ \ \ \ \ \ $ /*(2/3)
10/3=c

Now, remember that the triangle ABC can also be obtuse.
Thus the point C lies on (19/3|1) or (-1/3|1). (You either subtract 10/3 from 3 or add it).